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Bent Functions of maximal degree
Ayça Çeşmelioğlu and Wilfried Meidl
Abstract—In this article a technique for constructing p-ary
bent functions from plateaued functions is presented. This
generalizes earlier techniques of constructing bent from nearbent functions. The Fourier spectrum of quadratic monomials is
analysed, examples of quadratic functions with highest possible
absolute values in their Fourier spectrum are given. Applying
the construction of bent functions to the latter class of functions
yields bent functions attaining upper bounds for the algebraic
degree when p = 3, 5. Until now no construction of bent functions
attaining these bounds was known.
Index Terms—Bent functions, Fourier transform, algebraic
degree, quadratic functions, plateaued functions
I. I NTRODUCTION
Let p be a prime, and let Vn be any n-dimensional vector
space over Fp and f be a function from Vn to Fp . If p = 2
we call f a binary or Boolean function, if p is an odd prime
we call f a p-ary function. The Fourier transform of f is the
complex valued function fb on Vn given by
X
fb(b) =
fp (x)−hb,xi
x∈Vn
2πi/p
where p = e
and h, i denotes any inner product on Vn .
The function f is called a bent function if |fb(b)|2 = pn for
all b ∈ Vn . Whereas for p = 2, when p = −1 thus fb(b) is
an integer, bent functions can only exist for even n, for odd
p bent functions exist for both odd and even n, see [7].
The normalized Fourier coefficient at b ∈ Vn of a function
from Vn to Fp is defined by p−n/2 fb(b). A binary bent function
clearly must have normalized Fourier coefficients ±1, and for
the p-ary case we always have (cf. [5], [7, Property 8])
(
f ∗ (b)
±p
: n even or n odd, p ≡ 1 mod 4
−n/2 b
p
f (b) =
f ∗ (b)
±ip
: n odd and p ≡ 3 mod 4
(1)
where f ∗ is a function from Vn to Fp that by definition gives
the exponent of p .
A bent function f is called regular if
p−n/2 fb(b) = fp
∗
(b)
f ∗ (b)
for all b ∈ Vn , i.e., the coefficient of p
is always +1.
Observe that for a binary bent function this holds trivially. As
easily seen from (1) a p-ary regular bent function can only
exist for even n or for odd n when p ≡ 1 mod 4.
A bent function f is called weakly regular if, for all b ∈ Vn ,
we have
∗
p−n/2 fb(b) = ζ pf (b) .
The authors are with Sabancı University, MDBF, Orhanlı,
Tuzla, 34956 İstanbul, Turkey (email: [email protected],
[email protected]). The article was written when A. Ceşmelioğlu was
working at INRIA Paris-Rocquencourt.
for some complex number ζ with absolute value 1 (see [7]).
By (1), ζ can only be ±1 or ±i.
A function f from Vn to Fp is called plateaued if |fb(b)|2 =
A or 0 for all b ∈ Vn . Using (the special case of) Parseval’s
identity
2
X fb(b) = p2n
b∈Vn
n+s
we see that A = p
for an integer s with 0 ≤ s ≤ n.
We will call a plateaued function with |fb(b)|2 = pn+s or 0
an s-plateaued function. The case s = 0 corresponds to bent
functions by definition, and we have s = n if and only if f is
an affine function or constant. We remark that for 1-plateaued
functions the term near-bent function was used in [1], [8],
binary 1-plateaued and 2-plateaued functions are referred to
as semi-bent functions in [2].
It is well known that the maximal (algebraic) degree of a
binary bent function in dimension n is n/2 (see [10]). For
p-ary bent functions, p > 2, Hou [6] showed the following
bounds:
If f is a bent function from Vn to Fp then the degree deg(f )
of f satisfies
(p − 1)n
deg(f ) ≤
+ 1.
(2)
2
If f is weakly regular, then
(p − 1)n
.
(3)
2
As remarked in [6] p-ary Maiorana-McFarland bent functions
f from Fnp to Fp , which are always regular and for which n
is always even (cf [7]), can be used to attain the bound (3).
But in [6] it is left as an open problem if
I the bound (2) can be attained when n > 1,
II the bound (3) can be attained when n ≥ 3 is odd.
Only very recently a bent function from F27 to F3 found by
computer search attaining the bound (2) has been presented in
[11]. To our best knowledge no general construction of bent
functions attaining the bounds (2), (3) is known by now.
In [2], [8], [1] a method of constructing bent from nearbent functions has been presented. Applying this method, in
[8] the first examples of non-weakly-normal bent functions
(see [4]) in dimensions 10 and 12 have been presented, in
[1] the first known infinite classes of non-weakly regular bent
functions for arbitrary odd prime p have been introduced
(until then only sporadic ternary examples were known, and
a recursive method of obtaining an infinite family of nonweakly regular bent functions starting from one non-weakly
regular bent function, see [11]).
In this article we further develop the method of [2], [8],
[1], and obtain bent functions from s-plateaued functions.
In Section II we give some results on quadratic s-plateaued
deg(f ) ≤
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functions. In Section III we present the construction of bent
functions from s-plateaued functions. In Section IV we utilize
this construction to obtain bent functions of maximal degree.
In particular we construct p-ary bent functions for p = 3
attaining the upper bound (2), and p-ary bent functions for
p = 3, 5 in odd dimension that attain the upper bound (3),
which partly solves open problems I and II.
II. P RELIMINARIES
As all vector spaces of dimension n over Fp are isomorphic
we may associate Vn with the finite field Fpn . We then usually
use the inner product hx, yi = Trn (xy) where Trn (z) denotes
the absolute trace of z ∈ Fpn . In this framework the Fourier
transform of a function f from Fpn to Fp is the complex
valued function on Fpn given by
X
fp (x)−Trn (bx) .
fb(b) =
x∈Fpn
Recall that a function f from Fpn to Fp of the form
!
l
X
i
f (x) = Trn
ai xp +1
(4)
i=0
is called quadratic, its algebraic degree is two (if f is not
constant), see [2], [5]. As well known, every quadratic function
from Fpn to Fp is plateaued. The value for s can be obtained
with the standard squaring technique (see [1, Theorem 2], [5]):
X
|fb(−b)|2 =
pf (x)−f (y)+Trn (b(x−y))
or equivalently (see [9, p.118])
deg(gcd(A(x), xn − 1)) = s,
where
A(x) =
X
fp (z)+Trn (bz)
z∈Fpn
X
pf (y+z)−f (y)−f (z) .
y∈Fpn
Straightforward one gets f (y + z) − f (y) − f (z) =
!
l X
l
pl pl+i
pl−i pl−i
pl
Trn y
ai z
+ ai z
= Trn (y p L(z)).
l X
l
api xl+i + api
l−i
xl−i
(7)
i=0
is the associate of L(x).
In some sense the simplest quadratic functions are quadratic
monomials. It is well known that the monomial f (x) =
r
Trn (axp +1 ) is bent for every a ∈ F∗pn if and only if
n/ gcd(r, n) is odd ([3]). In [5] it was examined for which
r
a ∈ F∗pn the function f (x) = Trn (axp +1 ) is bent covering also the case when n/ gcd(r, n) is even. In [1] it was
r
shown that f (x) = Trn (axp +1 ) is never 1-plateaued. A full
treatment of quadratic monomials is given in the following
theorem. In particular we will see that quadratic monomials are
never s-plateaued for any odd s. At some positions in the proof
we will use that for a divisor s of n we have ps −1|(pn −1)/2
if and only if n/s is even, or equivalently ν(s) < ν(n) where
ν denotes the 2-adic valuation on integers.
Theorem 1: The
quadratic
monomial
f (x)
=
r
Trn (axp +1 ) ∈ Fpn [x] is s-plateaued for some a ∈ F∗pn
if and only if n is even, s is an even divisor of n and
ν(s) = ν(r) + 1.
Proof: The linearized polynomial L(x) corresponding to
r
f (x) = Trn (axp +1 ) ∈ Fpn [x] is given by
r
2r
L(x) = ax + ap xp .
x,y∈Fpn
=
(6)
We want to find out under which conditions ker(L) has
dimension s ≥ 2.
For a primitive element γ of Fpn , let a = γ c for some
c, 0 ≤ c ≤ pn − 2. Then L(γ t ) = 0 for an exponent
t, 0 ≤ t ≤ pn − 2, if and only if
i=0
γ
pn −1
−c(pr −1)
2
= γ (p
2r
−1)t
,
Consequently
|fb(−b)|2
=
X
fp (z)+Trn (bz)
z∈Fpn
X
= pn
X
which is equivalent to
pTrn (yL(z))
pn − 1
− c(pr − 1) ≡ (p2r − 1)t mod (pn − 1).
2
y pl ∈Fpn
fp (z)+Trn (bz)
The kernel has dimension s if and only if this congruence has
ps − 1 incongruent solutions, i.e.
z∈Fpn
L(z)=0
=
pn+s
0
if f (z) + Trn (bz) ≡ 0 on ker(L)
otherwise
where in the last step we used that f (z) + Trn (bz) is linear on
the kernel ker(L) of L. Summarizing, the square of the Fourier
transform of the quadratic function f in (4) takes absolute
values 0 and pn+s , where s is the dimension of the kernel of
the linear transformation on Fpn defined by
L(x) =
l X
l
api xp
l+i
+ api
l−i
xp
l−i
i=0
.
(5)
(a) gcd(p2r − 1, pn − 1) = pgcd(2r,n) − 1 = ps − 1, or
equivalently
gcd(2r, n)
= s,
n
(b) ps − 1| p 2−1 − c(pr − 1).
First, assume that n is odd. By (a) this implies that s is also odd
and hence gcd(r, n) = s. Consequently
ps − 1 divides pr − 1,
n
p
−1
thus (b) holds if and only if ps −1| 2 , which contradicts that
n is odd. Therefore for the rest of the proof we may assume
that n is even and hence by condition (a) also s is even. We
consider two cases.
•
Clearly this corresponds to
n
deg(gcd(L(x), xp − x)) = ps ,
Case I: ns is even.
Condition (a) then implies that 2r
s is odd, hence ν(s) =
ν(r) + 1. We remark that in this case and ps − 1 does
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not divide pr − 1. Due to condition (b) we then have to
find solutions c, y for the equation
pn − 1
.
(8)
y(ps − 1) + c(pr − 1) =
2
Equation (8) has solutions
if and only if gcd(ps − 1, pr −
pn −1
gcd(r,s)
1) = p
− 1 | 2 , which is guaranteed since n/s
is even.
n
• Case II: s is odd.
We consider two subcases:
(i) Suppose 2r
s is odd. Then we have ν(s) = ν(r) + 1
and hence ps −1 - pr −1. Condition (b) is satisfied for
some integer c if and only if equation (8) has solutions. This is guaranteed by ν(n) = ν(s) = ν(r)+1.
s
(ii) Suppose 2r
1|
s is even, i.e. ν(s) ≤ ν(r). Then p −
n
r
s
p − 1 and for condition (b), we need p − 1| p 2−1
which contradicts that n/s is odd.
2
We remark that for n, r, s satisfying the conditions of the
r
theorem, the elements a = γ c for which f (x) = Trn (axp +1 )
is s-plateaued are obtained from the congruence (8). For the
remaining elements a ∈ F∗pn the corresponding monomial is
bent.
III. O BTAINING BENT FUNCTIONS FROM s- PLATEAUED
FUNCTIONS
Let f be a function from Fpn to Fp , and fb denote its Fourier
transform. The support of fb is then defined to be the set
supp(fb) = {b ∈ Fpn | fb(b) 6= 0}. In this section we describe a
procedure to construct p-ary bent functions in dimension n+s
from s-plateaued functions from Fpn to Fp . The s-plateaued
functions must be chosen so that the supports of their Fourier
transforms are pairwise disjoint. Our construction can be seen
as a generalization of the constructions in [2], [1], [8] where
s = 1.
Theorem 2: For each a = (a1 , a2 , · · · , as ) ∈ Fsp , let fa (x)
be an s-plateaued function from Fpn to Fp . If supp(fba ) ∩
s
supp(fc
b ) = ∅ for a, b ∈ Fp , a 6= b, then the function
F (x, y1 , y2 , · · · , ys ) from Fpn × Fsp to Fp defined by
F (x, y1 , y2 , · · · , ys ) =
X (−1)s Qs yi (yi − 1) · · · (yi − (p − 1))
i=1
(y1 − a1 ) · · · (ys − as )
a∈Fsp
fa (x)
is bent.
Proof: For (α, a), (x, y) ∈ Fpn × Fsp the inner product we use
is Trn (αx)+a·y = Trn (αx)+a1 y1 +a2 y2 +· · ·+as ys , where
a = (a1 , · · · , as ), y = (y1 , · · · , ys ). The Fourier transform Fb
of F at (α, a) is
X
(x,y1 ,···,ys )−Trn (αx)−a·y
Fb(α, a) =
F
p
x∈Fpn ,y1 ,···,ys ∈Fp
=
X
p−a·y
y1 ,···,ys ∈Fp
=
X
y1 ,···,ys ∈Fp
=
X
y1 ,···,ys ∈Fp
p−a·y
X
pF (x,y1 ,···,ys )−Trn (αx)
As each α ∈ Fpn belongs to the support
of exactly one fc
y,
n+s
b
−a·y c
s
y ∈ Fp , for this y we have F (α, a) = |p fy (α)| = p 2 .
2
Theorem 3: For each a ∈ Fsp , let ga be a quadratic splateaued function from Fpn to Fp with the corresponding
linearized polynomial La such that for all a ∈ Fsp , La has the
same kernel {c1 β1 + · · · + cs βs , 0 ≤ c1 , · · · , cs ≤ p − 1} in
Fpn . For each a = (a1 , · · · , as ) ∈ Fsp , let γa ∈ Fpn be such
that
ga (βj ) + Trn (γa βj ) = g0 (βj ) + aj ,
(9)
for all j = 1, · · · , s. Then for each a ∈ Fsp , the s-plateaued
function fa defined by fa (x) = ga (x) + Trn (γa x) satisfies
s
supp(fba ) ∩ supp(fc
b ) = ∅ for b ∈ Fp , a 6= b.
Proof: We have to show that −α ∈ supp(fc
b ) implies −α 6∈
supp(fba ) for a 6= b. Suppose −α ∈ supp(fc
b ), i.e.
gb (βj )+Trn (γb βj )+Trn (αβj ) = g0 (βj )+bj +Trn (αβj ) = 0,
for each j = 1, · · · , s.
Let a 6= b and suppose that aj 6= bj for some 1 ≤ j ≤ s.
Then
fa (βj ) + Trn (αβj ) = ga (βj ) + Trn (γa βj ) + Trn (αβj ) =
g0 (βj ) + aj + Trn (αβj ) 6= 0.
2
Observe that the existence of γa that satisfies equation (9)
for all j = 1, . . . , s, is guaranteed by the linear independence
of β1 , . . . , βs , the elements of the basis of the kernel for the
linearized polynomial La . To point to a deterministic way of
obtaining γa we give a detailed argument for this observation:
Let {δ1 , δ2 , . . . , δn } and {ρ1 , ρ2 , . . . , ρn } be dual bases of Fpn
over Fp , and let βj = bj1 δ1 + bj2 δ2 + · · · + bjn δn . Put γa =
x1 ρ1 + x2 ρ2 + · · · + xn ρn , then
Trn (γa βj ) = bj1 x1 + bj2 x2 + · · · + bjn xn .
A value for γa is then a solution of the linear system
Bx = c over Fp for the (s × n)-matrix B = (bjk )
and c = (c1 , · · · , cs )T with cj = g0 (βj ) + aj − ga (βj ),
j = 1, 2, . . . , s.
Example 1: To obtain a 2-plateaued monomial function
r
Tr4 (ax3 +1 ) from F34 to F3 , by Theorem 1 we can choose
r = 1 or r = 3. In fact the monomials g0 (x) =
Tr4 (x4 ), g1 (x) = Tr4 (x28 ) have the same corresponding
2
linearized polynomial L(x) = x + x3 with a kernel of
dimension 2 in F34 . A basis for this kernel is {β, β 3 } where
β is a root of the polynomial x4 + x2 + 2. Since we
have g0 (β) = g0 (β 3 ) = g1 (β) = g1 (β 3 ) = 0 for each
a = (a1 , a2 ) ∈ F3 × F3 , we choose γa ∈ F34 such that
Tr4 (γa β) = a1 , Tr4 (γa β 3 ) = a2 . For instance we can choose
x∈Fpn
γ(0,0) = 1,
γ(0,1) = β 3 + 1,
γ(0,2) = 2β 3 + 2,
X
γ(1,0) = β,
γ(1,1) = β 3 + β,
γ(1,2) = 2β 3 + β + 1,
γ(2,0) = 2β + 1,
γ(2,1) = β 3 + 2β,
γ(2,2) = 2β 3 + 2β,
pfy (x)−Trn (αx)
x∈Fn
p
p−a·y fc
y (α).
and for the nine required 2-plateaued functions with pairwise
disjoint supports of their Fourier transforms we then can
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choose
f(0,0) (x) = Tr4 (x4 + x),
f(0,1) (x) = Tr4 (x4 + (β 3 + 1)x),
f(0,2) (x) = Tr4 (x4 + (2β 3 + 2)x),
f(1,0) (x) = Tr4 (x4 + βx),
f(1,1) (x) = Tr4 (x4 + (β 3 + β)x),
4
is (n − 1)-plateaued.
Proof: We only show the statement for n even, the case where
n is odd is shown in the same way. Straightforward one sees
that the associate A(x) (7) of the linearized polynomial (5)
that corresponds to ch(x) is given by
A(x)
=
f(1,2) (x) = Tr4 (x28 + (2β 3 + β + 1)x),
f(2,0) (x) = Tr4 (x28 + (2β + 1)x),
f(2,1) (x) = Tr4 (x28 + (β 3 + 2β)x),
p + 1 n/2
p+1 n p+1
x +
+2
x
+
2
2
2
!
n/2−1
X
xn/2+i + xn/2−i
c
i=1
f(2,2) (x) = Tr4 (x28 + (2β 3 + 2β)x).
n−1
The function
F (x, y, z) =
X
a=(a1 ,a2 )
∈F3 ×F3
yz(y − 1)(z − 1)(y − 2)(z − 2)
fa (x)
(y − a1 )(z − a2 )
is then bent.
IV. B ENT FUNCTIONS WITH HIGH ALGEBRAIC DEGREE
In this section we construct 3-ary bent functions attaining
the bound (2), and 3-ary and 5-ary bent functions attaining
the bound (3) also for odd dimension. This can be seen as the
main result of this paper. We start with a proposition on the
degree of the bent functions constructed in Theorem 2 with
quadratic s-plateaued functions.
Proposition 1: Let {fa | a ∈ Fsp } be a set of quadratic splateaued
functions satisfying the conditions of Theorem 2.
P
If a∈Fsp fa is quadratic (affine) then the bent function F in
Theorem 2 has degree deg(F ) = (p − 1)s + 2 (deg(F ) =
(p − 1)s + 1).
P
Proof: If the quadratic terms in
a∈Fsp fa do not cancel, then in F given P
as in Theorem 2 the summand
(−1)s y1p−1 y2p−1 · · · ysp−1 a∈Fsp fa (x) having degree (p −
P
1)s + 2 does not vanish. Similarly, if a∈Fsp fa = Trn (cx)
for some c ∈ Fpn , then the bent function F in Theorem 2
has the summand (−1)s y1p−1 y2p−1 · · · ysp−1 Trn (cx) of degree
deg(F ) = (p − 1)s + 1 as term of largest degree.
2
In order to obtain bent functions of highest possible degree
we have to choose s as large as possible. Naturally we have
s ≤ n and the set of n-plateaued functions precisely coincides
with the set of affine and constant functions. Thus s = n − 1
is the maximal value for s for s-plateaued quadratic functions.
The following proposition shows that this maximal value can
be obtained.
Proposition 2: For n even, the quadratic function ch(x),
c ∈ F∗p , from Fpn to Fp with


n/2−1
X
n/2
i
p
+
1
p
+
1
xp +1 +
x2 +
xp +1 
h(x) = Trn 
2
2
i=1
(10)
is (n − 1)-plateaued.
If n is odd, then the quadratic function ch(x), c ∈ F∗p , from
Fpn to Fp with


(n−1)/2
X
i
p+1 2
h(x) = Trn 
x +
xp +1 
(11)
2
i=1
p+1 n p+1 X i
x −
+
x
2
2
i=0
=
c
=
n−1
X
p+1 n
c
(x − 1) + c
xi .
2
i=0
Evidently we have gcd(A(x), xn − 1) =
is (n − 1)-plateaued by equation (6).
Pn−1
i=0
!
xi , thus h(x)
2
For the subsequent theorem we fix the following notation:
1) h is the function (10) and (11) when n is even and odd,
respectively,
let
2) for each a = (a1 , a2 , . . . , an−1 ) ∈ Fn−1
p
• ca be a nonzero element of Fp ,
• γa be an element of Fpn , such that
ca h(βj )+Trn (γa βj ) = c0 h(βj )+aj , j = 1, . . . , n−1,
(12)
for a fixed basis {β1 , β2 , . . . , βn−1 } for the kernel
of the linearized polynomial L corresponding to h,
• ha (x) = ca h(x) + Trn (γa x).
Theorem 4: The function F defined by
F (x, y1 , y2 , · · · , yn−1 ) =
X (−1)n−1 Qn−1 yi (yi − 1) · · · (yi − (p − 1))
i=1
ha (x)
(y1 − a1 ) · · · (yn−1 − an−1 )
n−1
a∈Fp
from Fpn × Fn−1
to P
Fp is bent, and has degree
p
(p − 1)(n − 1) + 2 if
ca 6= 0, and degree
a∈Fn−1
p
P
P
(p − 1)(n − 1) + 1 if a∈Fn−1
c
=
0
and a∈Fn−1
γa 6= 0.
a
p
p
n−1
Proof: For all a ∈ Fp
the linearized polynomials La
corresponding to ca h have the same kernel, and the definition
(12) for γa guarantees that the Fourier transforms of the
(n − 1)-plateaued functions ha , a ∈ Fn−1
, have pairwise
p
disjoint
support.
By
Theorem
2
the
function
F
P
P
P is bent. Since
n−1 ha
n−1 ca )h + Trn (
n−1 γa x) has
=
(
a∈Fp
a∈Fp
a∈F
P
Pp
degree 2 if a∈Fn−1
c
=
6
0
and
degree
1
if
ca = 0
a
a∈Fn−1
p
p
P
n−1 γa
and
=
6
0,
the
bent
function
F
has
degree
a∈Fp
(p − 1)(n − 1) + 2 and (p − 1)(n − 1) + 1, respectively, by
Proposition 1.
2
P
Of course it is always possible to choose a∈Fn−1
ca 6= 0.
p
We emphasize that for any choice of the setP
{ca | a ∈ Fpn−1 }
we can choose {γa | a ∈ Fn−1
} such that a∈Fn−1
γa 6= 0,
p
p
since for every a ∈ Fpn−1 the linear system from which we
obtain γa does not have a unique solution.
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Considering that the bent function F in Theorem 4 is in
dimension n + s = 2n − 1 the bounds (2) and (3) become
deg(f ) ≤ (p − 1)(n − 1) + (p − 1)/2 + 1 and deg(f ) ≤
(p − 1)(n − 1) + (p − 1)/2, respectively. As we can see the
function
P F in Theorem 4 attains the first bound when p = 3
and a∈Fn−1
ca 6= 0, and the second bound for arbitrary odd
p
P
P
dimensions when p = 3, a∈Fn−1
ca = 0, a∈Fpn−1 γa 6= 0,
p
P
and when p = 5, a∈Fn−1
ca 6= 0. We obtained the following
p
corollaries partially solving open problems I and II.
Corollary 1: The bounds (2) and (3) can be attained for
p = 3 in arbitrary odd dimension.
Corollary 2: The bound (3) can be attained for p = 5 in
arbitrary odd dimension.
Remark 1: As the bound (2) can only be attained by nonweakly regular bent functions, the function P
F in Theorem 4
must be non-weakly regular for p = 3 and a∈Fn−1
ca 6= 0.
p
In fact this can be seen by a generalization of the arguments in
[1] where infinite classes of non-weakly regular bent functions
have been constructed. The reason behind F being non-weakly
regular is that we cannot choose all ca with
P the same quadratic
character under the condition p = 3 and a∈Fn−1
ca 6= 0. This
p
is not a problem for p = 5 where non-weakly regular as well
as weakly regular F attaining the bound (3) can be found. We
recall that for weakly regular bent functions the bound (3) is
best possible.
V. C ONCLUSIONS
In [1], [2], [8] a construction of bent functions from nearbent functions has been presented. In this article we generalize
the this construction and present a technique to construct bent
functions from plateaued functions. As quadratic functions are
always plateaued we investigate some classes of quadratic
functions. We completely describe the Fourier spectrum of
quadratic monomials, and present classes of quadratic functions with maximal possible absolute values in their Fourier
spectrum. We apply our construction to the latter classes of
quadratic functions and thereby obtain the first construction
of bent functions in characteristic 3 and 5 attaining upper
bounds on the algebraic degree of bent functions presented by
Hou in [6]. This partly solves open problems on the degree of
bent functions and we can reformulate the open problems as
follows:
(i) Can the bound (2) be attained for p ≥ 5;
(ii) Can the bound (2) be attained in even dimension;
(iii) Can the bound (3) be attained for p ≥ 7 and odd
dimension.
ACKNOWLEDGMENTS
The first author would like to thank the hospitality of INRIA
Paris-Rocquencourt and she also wishes to thank Prof. Pascale
Charpin who made the research visit possible.
Parts of the article were written during a short visit of the
second author at INRIA Paris-Rocquencourt. He would like
to thank the institute for the hospitality.
5
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Ayça Çeşmelioğlu received the Ph.D. degree in mathematics from Sabancı
University, İstanbul, Turkey, in June 2008.
Currently, she is a postdoc researcher at Sabanci University. Her research
interests include permutation polynomials, cryptographically significant functions and pseudorandom sequences.
Meidl Wilfried received the M.Sc and Ph.D degrees from Klagenfurt
University, Austria, in 1994 and 1998, respectively.
From 2000 to 2002, ha was with the Institute of Discrete Mathematics,
OEAW, Vienna, Austria. From 2002 to 2004, ha was with Temasek Labs,
National University of Singapore and from 2004 to 2005 with RICAM, Linz,
Austria. He is now with Sabancı University, MDBF, İstanbul, Turkey. His
research interests include sequences, permutation polynomials, pseudorandom
number generation, finite fields and their applications, APN-functions, bentfunctions.